Introduction to Acids and Bases - AP Chemistry

1. In the reaction $\mathrm{HCl(aq) + H_2O(l) \rightarrow H_3O^+(aq) + Cl^-(aq)}$, which species is the conjugate base of the acid?
   Answer: $\mathrm{Cl^-}$ This question tests introduction to acids and bases focusing on conjugate base identification in the Brønsted-Lowry framework. In the reaction HCl + H₂O → H₃O⁺ + Cl⁻, we need to find what remains after the acid donates its proton. HCl donates an H⁺ to water, leaving behind Cl⁻, while water accepts that H⁺ to become H₃O⁺. The conjugate base of an acid is what remains after proton donation, so Cl⁻ is the conjugate base of HCl. A common mistake is choosing H₃O⁺ (choice A) because students associate it with acids, but H₃O⁺ is actually the conjugate acid of water in this reaction. To find a conjugate base, identify the acid first, then determine what species remains after it loses one H⁺.
2. In the reaction $\mathrm{H_2PO_4^-(aq) + H_2O(l) \rightleftharpoons HPO_4^{2-}(aq) + H_3O^+(aq)}$, which species is acting as the Brønsted–Lowry base?
   Answer: $\mathrm{H_2O}$ This question tests introduction to acids and bases using Brønsted-Lowry definitions to identify the base in a reaction. In the reaction H₂PO₄⁻ + H₂O ⇌ HPO₄²⁻ + H₃O⁺, we must track which species accepts a proton. H₂PO₄⁻ loses an H⁺ to become HPO₄²⁻ (acting as an acid), while H₂O gains that H⁺ to become H₃O⁺ (acting as a base). Therefore, water (H₂O) is the Brønsted-Lowry base in this reaction. Students might incorrectly choose HPO₄²⁻ (choice B) thinking the product with more negative charge is the base, but HPO₄²⁻ is the conjugate base of H₂PO₄⁻, not the base in this reaction. To identify the base, find the species that gains H⁺ during the reaction process.
3. A student adds solid sodium bicarbonate to water and observes that the solution can react with both added HCl(aq) and added NaOH(aq). Which statement best describes the acid–base behavior of the bicarbonate ion, $\mathrm{HCO_3^-}$, in water?
   Answer: $\mathrm{HCO_3^-}$ is amphiprotic; it can both donate and accept a proton. This question tests introduction to acids and bases using the Brønsted-Lowry definition and the concept of amphiprotic species. The bicarbonate ion HCO₃⁻ can act as both an acid (donating H⁺ to form CO₃²⁻) and a base (accepting H⁺ to form H₂CO₃), which is why it reacts with both HCl (acting as a base) and NaOH (acting as an acid). This dual behavior defines an amphiprotic substance - one that can both donate and accept protons. Choice B is incorrect because having hydrogen doesn't automatically make something only an acid; HCO₃⁻ can also accept protons. The key strategy is to identify species that have both removable hydrogen atoms AND the ability to accept protons, making them amphiprotic.
4. A student writes the reaction $\mathrm{HCO_3^-(aq) + H_2O(l) \rightleftharpoons H_2CO_3(aq) + OH^-(aq)}$. In this reaction, $\mathrm{HCO_3^-}$ is best classified as which of the following?
   Answer: A Brønsted–Lowry base (proton acceptor) This question tests introduction to acids and bases using the Brønsted-Lowry definition to classify species behavior. In the reaction HCO₃⁻(aq) + H₂O(l) ⇌ H₂CO₃(aq) + OH⁻(aq), we track proton movement to identify acids and bases. HCO₃⁻ gains an H⁺ to become H₂CO₃, which means HCO₃⁻ accepts a proton, making it a Brønsted-Lowry base. The H⁺ comes from H₂O, which loses it to become OH⁻, confirming water acts as the acid here. Students might choose option A thinking HCO₃⁻ donates a proton because it contains hydrogen, but in this specific reaction it's accepting, not donating. The key is to compare the reactant form (HCO₃⁻) with its product form (H₂CO₃) to see whether H⁺ was gained (base) or lost (acid).
5. A student writes the proton-transfer reaction $\mathrm{HNO_2(aq) + OH^-(aq) \rightarrow NO_2^-(aq) + H_2O(l)}$. Which species is the Brønsted–Lowry acid?
   Answer: $\mathrm{HNO_2}$ This question assesses the skill of introduction to acids and bases, focusing on the Brønsted–Lowry acid as the proton donor. In the reaction HNO₂(aq) + OH⁻(aq) → NO₂⁻(aq) + H₂O(l), proton transfer occurs from HNO₂ to OH⁻. HNO₂ loses H⁺ to become NO₂⁻, confirming it as the acid. This determines choice D as correct. A tempting distractor is B, OH⁻, but it accepts the proton, acting as the base, not the acid. Always track H⁺ transfer; the acid is the species that donates the proton in the reaction.
6. Which species can act as a Brønsted–Lowry base in water because it can accept a proton?

   Species	Formula	Feature
   Hydronium	$\mathrm{H_3O^+}$	has an extra proton
   Water	$\mathrm{H_2O}$	has lone pairs on O
   Nitrate	$\mathrm{NO_3^-}$	resonance-stabilized anion

   Answer: $\mathrm{H_2O}$ and $\mathrm{NO_3^-}$ This question assesses the skill of introduction to acids and bases, using the Brønsted–Lowry definition of bases as proton acceptors. Among the species, H₂O and NO₃⁻ can accept H⁺: H₂O to H₃O⁺ via lone pairs, NO₃⁻ to HNO₃ due to its charge. H₃O⁺, however, donates H⁺ rather than accepts. This selects choice E as correct. A tempting distractor is D, including H₃O⁺ and H₂O, but H₃O⁺ is an acid, not a base. Always track H⁺ transfer; bases have features like lone pairs or negative charges that enable proton acceptance.
7. When ammonia is dissolved in water, the following reaction occurs: $\mathrm{NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)}$. In this reaction, which species is the Brønsted–Lowry acid?
   Answer: $\mathrm{H_2O}$ This question tests introduction to acids and bases using the Brønsted-Lowry definition of acids as proton donors. In the reaction NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq), we need to track which species loses H⁺. Water (H₂O) donates a proton to ammonia, becoming OH⁻, while NH₃ accepts that proton to become NH₄⁺. Therefore, H₂O is the Brønsted-Lowry acid and NH₃ is the base. Choice C (NH₄⁺) is tempting but incorrect because it's a product, not a reactant, and it's the conjugate acid of NH₃, not the acid in this reaction. To identify acids in reactions, look for the species that loses H⁺ going from reactants to products.
8. Consider the reaction $\mathrm{HNO_2(aq) + OH^-(aq) \rightarrow NO_2^-(aq) + H_2O(l)}$. Which statement correctly identifies the Brønsted–Lowry acid and base (forward direction)?
   Answer: $\mathrm{HNO_2}$ is the acid and $\mathrm{OH^-}$ is the base. This question tests the skill of introduction to acids and bases, using Brønsted–Lowry definitions to identify acid as proton donor and base as acceptor. In the forward reaction, HNO2 loses H+ to OH-, forming NO2- and H2O. OH- accepts the H+, confirming it as the base. Thus, HNO2 is the acid and OH- is the base. A tempting distractor is choice A, but it reverses the roles incorrectly. Always track H+ transfer direction; donor is acid, acceptor is base in the specified direction.
9. A student observes the reaction $\mathrm{HS^-(aq) + H_2O(l) \rightleftharpoons H_2S(aq) + OH^-(aq)}$. In the forward direction, water is best classified as which of the following?
   Answer: A Brønsted–Lowry acid This question tests introduction to acids and bases by classifying water's role using Brønsted-Lowry definitions. In the reaction HS⁻(aq) + H₂O(l) ⇌ H₂S(aq) + OH⁻(aq), we track proton movement in the forward direction. HS⁻ gains an H⁺ to become H₂S, while H₂O loses an H⁺ to become OH⁻. Since water donates a proton to HS⁻, water acts as a Brønsted-Lowry acid in this reaction. Students might choose option B thinking water always acts as a base, but water is amphiprotic and its role depends on what it's reacting with. The strategy is to check whether water gains H⁺ (base) or loses H⁺ (acid) by comparing H₂O to its product form.
10. A student mixes aqueous acetic acid and water: $\mathrm{CH_3COOH(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + CH_3COO^-(aq)}$. Which pair is a conjugate acid–base pair?
    Answer: $\mathrm{CH_3COOH}$ and $\mathrm{CH_3COO^-}$ This question tests the skill of introduction to acids and bases, emphasizing conjugate pairs where the acid and its conjugate base differ by one proton. In the reaction, identify pairs by comparing formulas before and after proton transfer. CH3COOH donates H+ to become CH3COO-, making them a conjugate acid-base pair. H3O+ and H2O form another pair, but the question asks for one specific pair. A tempting distractor is CH3COOH and H3O+ (choice C), but they are both acids, not conjugates. To find conjugate pairs, compare species differing by H+; acid becomes its conjugate base after donation.
11. A student dissolves sodium acetate in water, and the following reaction occurs: $\mathrm{CH_3COO^- (aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)}$. Which species acts as the Brønsted–Lowry base in this reaction?
    Answer: $\mathrm{CH_3COO^-}$ This question tests introduction to acids and bases using the Brønsted-Lowry definition, where bases accept protons (H⁺). In the reaction CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, we need to track which species gains a proton. CH₃COO⁻ (acetate ion) gains an H⁺ to become CH₃COOH (acetic acid), while H₂O loses an H⁺ to become OH⁻. Since CH₃COO⁻ accepts a proton, it acts as the Brønsted-Lowry base. A common mistake is choosing H₂O (choice A) because students might think water is always a base, but here water donates a proton and acts as an acid. To identify Brønsted-Lowry bases, always look for the species that accepts H⁺ in the reaction.
12. A student examines the following reaction in aqueous solution: $\mathrm{CN^-(aq) + H_2O(l) \rightleftharpoons HCN(aq) + OH^-(aq)}$. Which statement correctly identifies the conjugate acid of $\mathrm{CN^-}$?
    Answer: $\mathrm{HCN}$ This question tests introduction to acids and bases focusing on identifying conjugate acid relationships. In the reaction CN⁻(aq) + H₂O(l) ⇌ HCN(aq) + OH⁻(aq), the cyanide ion CN⁻ accepts a proton from water to become HCN. The conjugate acid of any base is what forms when the base gains one proton. Since CN⁻ gains H⁺ to form HCN, HCN is the conjugate acid of CN⁻. Choice D (CN⁻) is incorrect because a species cannot be its own conjugate acid - conjugate pairs must differ by one H⁺. To find a conjugate acid, add one H⁺ to the base; to find a conjugate base, remove one H⁺ from the acid.
13. In water, ammonia reacts according to $\mathrm{NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)}$. Which species acts as the Brønsted–Lowry acid in the forward reaction?
    Answer: $\mathrm{H_2O}$ This question tests the skill of introduction to acids and bases, focusing on the Brønsted–Lowry definition where an acid donates a proton and a base accepts one. In the given reaction, examine the forward direction to identify proton transfer. Water (H2O) loses a proton (H+) to ammonia (NH3), resulting in NH4+ and OH-. This proton donation confirms H2O as the Brønsted–Lowry acid. A tempting distractor is NH3 (choice A), but it is the base because it accepts the proton rather than donating it. Always track H+ transfer; the donor is the acid, and what remains is its conjugate base.
14. A student considers the reaction $\mathrm{H_3PO_4(aq) + H_2O(l) \rightleftharpoons H_2PO_4^-(aq) + H_3O^+(aq)}$. Which species is the conjugate acid of $\mathrm{H_2O}$ in this reaction?
    Answer: $\mathrm{H_3O^+}$ This question assesses the skill of introduction to acids and bases, emphasizing conjugate pairs where the conjugate acid gains a proton from the base. In the reaction H₃PO₄(aq) + H₂O(l) ⇌ H₂PO₄⁻(aq) + H₃O⁺(aq), proton transfer is from H₃PO₄ to H₂O. H₂O accepts H⁺ to become H₃O⁺, making H₃O⁺ the conjugate acid of H₂O. This selects choice B as correct. A tempting distractor is C, H₃PO₄, but it is the original acid, not related to H₂O's conjugate. Always track H⁺ transfer; the conjugate acid is the species formed when the base accepts a proton.
15. A student dissolves sodium acetate in water and considers the reaction $\mathrm{CH_3COO^-(aq) + H_2O(l) \rightleftharpoons CH_3COOH(aq) + OH^-(aq)}$. Which species is the Brønsted–Lowry acid in the forward reaction?
    Answer: $\mathrm{H_2O}$ This question tests introduction to acids and bases by identifying the Brønsted-Lowry acid in a hydrolysis reaction. In CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq), we need to identify which species donates a proton in the forward reaction. CH₃COO⁻ gains an H⁺ to become CH₃COOH (acting as base), while H₂O loses an H⁺ to become OH⁻, making water the proton donor and thus the Brønsted-Lowry acid. Students might incorrectly choose CH₃COOH (option A) because acetic acid is typically an acid, but in this reaction CH₃COOH is the product formed when the base accepts a proton. Remember to focus on the forward reaction and identify which reactant loses H⁺—that's your acid.
16. A student observes the reaction $\mathrm{H_2PO_4^-(aq) + H_2O(l) \rightleftharpoons HPO_4^{2-}(aq) + H_3O^+(aq)}$. In this reaction, $\mathrm{H_2PO_4^-}$ acts as
    Answer: a Brønsted–Lowry acid because it donates $\mathrm{H^+}$ This question assesses the skill of introduction to acids and bases, focusing on the Brønsted–Lowry acid as a proton donor. In the reaction H₂PO₄⁻(aq) + H₂O(l) ⇌ HPO₄²⁻(aq) + H₃O⁺(aq), proton transfer is from H₂PO₄⁻ to H₂O. H₂PO₄⁻ loses H⁺ to become HPO₄²⁻, acting as the acid. This identifies choice B as correct. A tempting distractor is A, calling it a base, but the reaction shows donation, not acceptance. Always track H⁺ transfer; the species donating H⁺ is the acid in the context of the reaction.
17. Consider the reaction in water: $\mathrm{CH_3COOH(aq) + H_2O(l) \rightleftharpoons CH_3COO^-(aq) + H_3O^+(aq)}$. Which species is the conjugate base of $\mathrm{CH_3COOH}$?
    Answer: $\mathrm{CH_3COO^-}$ This question tests introduction to acids and bases by identifying conjugate base relationships in Brønsted-Lowry theory. In the reaction CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq), the conjugate base of an acid is what remains after the acid donates its proton. CH₃COOH loses one H⁺ to become CH₃COO⁻, making CH₃COO⁻ the conjugate base of CH₃COOH. The proton is transferred to H₂O, which becomes H₃O⁺. Students might incorrectly choose H₂O (option B) thinking it's related to CH₃COOH, but H₂O is actually the base in this reaction that forms its own conjugate acid (H₃O⁺). Remember: to find a conjugate base, remove one H⁺ from the acid; to find a conjugate acid, add one H⁺ to the base.
18. A student mixes two aqueous solutions and the net ionic equation is $\mathrm{H^+(aq) + F^-(aq) \rightarrow HF(aq)}$. In this reaction, $\mathrm{F^-}$ is acting as
    Answer: a Brønsted–Lowry base because it accepts $\mathrm{H^+}$ This question assesses the skill of introduction to acids and bases, applying the Brønsted–Lowry definition where a base accepts a proton. In the reaction H⁺(aq) + F⁻(aq) → HF(aq), proton transfer is to F⁻ from H⁺. F⁻ accepts the H⁺ to form HF, making it the base. This determines choice B as correct. A tempting distractor is A, suggesting acid behavior, but no proton is donated by F⁻ here. Always track H⁺ transfer; the acceptor in the reaction is the base, forming its conjugate acid.
19. When hydrogen sulfide dissolves in water, one step can be represented as $\mathrm{H_2S(aq) + H_2O(l) \rightleftharpoons HS^-(aq) + H_3O^+(aq)}$. In this reaction, what is the Brønsted–Lowry acid?
    Answer: $\mathrm{H_2S}$ This question tests introduction to acids and bases using the Brønsted-Lowry definition, where acids donate protons (H⁺). In the reaction H₂S + H₂O ⇌ HS⁻ + H₃O⁺, we need to identify which species loses a proton. H₂S loses one H⁺ to become HS⁻, while H₂O gains that H⁺ to become H₃O⁺. Since H₂S donates a proton, it acts as the Brønsted-Lowry acid. A tempting distractor is H₃O⁺ (choice B) because students often associate H₃O⁺ with acids, but H₃O⁺ is the product formed when water accepts a proton, not the acid in this reaction. To identify Brønsted-Lowry acids, track which species loses H⁺ during the reaction.
20. A student mixes aqueous solutions of ammonia and hydrochloric acid, producing the reaction $\mathrm{NH_3(aq) + HCl(aq) \rightarrow NH_4^+(aq) + Cl^-(aq)}$. According to the Brønsted–Lowry definition, which species acts as the base in this reaction?
    Answer: $\mathrm{NH_3}$ This question tests introduction to acids and bases using the Brønsted-Lowry definition, where acids donate protons (H⁺) and bases accept protons. In the reaction NH₃(aq) + HCl(aq) → NH₄⁺(aq) + Cl⁻(aq), we need to track which species gains a proton. NH₃ starts without a positive charge and becomes NH₄⁺, meaning it gained an H⁺ ion (proton), making NH₃ the base. HCl loses its H⁺ to become Cl⁻, confirming HCl is the acid that donated the proton. A common mistake is choosing HCl (option D) because students might think "HCl is always an acid," but the question asks specifically for the base in this reaction. The key strategy is to identify which species gains H⁺ (base) versus which loses H⁺ (acid) by comparing reactants to products.

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